How do I evaluate $\int_{0}^{t} x^{\alpha+k-1}(t-x)^{\beta+k-1} dx$?

Question

Evaluate $\int_{0}^{t} x^{\alpha+k-1}(t-x)^{\beta+k-1} dx$

I find this very difficult to evaluate. Please help me. This is under the chapter of beta functions. But, I cannot see exactly where to use it. Please.

Answer 1

Let the original integral be $I$. Then putting $x=ty, \implies dx=t ~dy$, when $x=0,y=0;x=t,y=1$. So, $I=\int_{0}^{1} t^{\alpha+\beta+2k-1} ~y^{\alpha+k-1}(1-y)^{\beta+k-1} dy=t^{\alpha+\beta+2k-1} \int_{0}^{1} ~y^{\alpha+k-1}(1-y)^{\beta+k-1} dy=\boxed{~~t^{\alpha+\beta+2k-1}~\frac{\Gamma(\alpha+k)\Gamma(\beta+k)}{\Gamma(\alpha + \beta +2k)}~~}$

Answer 2

You know that the function beta is

$\beta(a,b):=\int_0^1 x^{a-1}(1-x)^{b-1}dx$

Then in your case if you consider the change of variables $y:=\frac{x}{t}$ you have that

$\int_0^t x^{\alpha+k-1}(t-x)^{\beta+k-1}dx= $

$t^{\alpha+\beta+2k-1}\int_0^1 y^{\alpha+k-1}(1-y)^{\beta+k-1}dy=$

$ =t^{\alpha+\beta+2k-1}\beta(\alpha+k,\beta+k) $